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Brent Hanneson

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Everything posted by Brent Hanneson

  1. 1,6k4 is a 4-digit number. So, 1^4 + 6^4 + k^4 + 4^4 = 16k4 Evaluate: 1 + 1296 + k^4 + 256 = 16k4 Simplify: 1553 + k^4 = 16k4 Whatever k is, it must be the case that the UNITS digit of k^4 is 1, so that 1553 + k^4 = 16k4 Test some values.. k = 1: 1^4 = 1, so we get: 1553 + 1 = 1614 NO GOOD k = 3: 3^4 = 81, so we get: 1553 + 81 = 1634 WORKS!! Answer: k = 3 Cheers, Brent
  2. A quick solution is to say that the original 20 measurements were all 34 That is {34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34} These values definitely have an average of 34 For security reasons the scientist coded the data by multiplying each of the measurements by 10 and then adding 40 to each product. 34 times 10 equals 340. Add 40 to get 380 So, the 20 coded values are {380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380, 380} New average = 380 Answer: B Cheers, Brent
  3. There's no math required for this question. The strip can be any value, since we could choose a certain width for the strip and then just shrink or expand the inlay until the ratio of (inlay area): (strip area) is 25:39. Cheers, Brent
  4. Another way to state this question is: What is the greatest common factor of 4000 and 180?
  5. G = 10^100 We can rewrite this as follows: G = (10^3)(10^97) Or we can say G = (1000)(10^97) So, G/8 = (1000)(10^97)/8 = (125)(10^97) G/5 = (1000)(10^97)/5 = (200)(10^97) G/4 = (1000)(10^97)/4 = (250)(10^97) G/2 = (1000)(10^97)/2 = (500)(10^97) So, G/8 + G/5 + G/4 + G/2 = (125)(10^97) + (200)(10^97) + (250)(10^97) + (500)(10^97) = (10^97)(125 + 200 + 250 + 500) = (10^97)(1075) This evaluates to be 1075 followed by 97 zeros So, the sum of the digits = 1 + 0 + 7 + 5 + a bunch of zeros = 13 Answer:
  6. We have an online study guide here - Study Guide - Overview | Greenlight Test Prep Online Course It can be used in conjunction with our free GRE video course. Cheers, Brent
  7. You will notice that I never said that 12 people can be seated at a round table in 11! ways (Chromepal made this claim). I'm not a big fan of "people sitting at a circle" questions, because the assumption is that the seats are identical. Since this is not stated in the question, I solved it in a way that does not require us to make this assumption. Cheers, Brent
  8. When it comes to remainders, we have a nice rule that says: If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc. When the positive integer n is divided by 3, the remainder is 2 So, some possible values of n are: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29,....etc When n is divided by 5, the remainder is 1. So, some possible values of n are: 1, 6, 11, 16, 21, 26, 31 .... etc We can see that 11 is the smallest number that BOTH lists share For more info on remainders, see our free video: GRE Integer Properties | Greenlight Test Prep Cheers, Brent
  9. 175 flashcards covering every formula, concept and strategy needed for the quantitative sections of the GRE. Each flashcard is linked to a corresponding video lesson from Greenlight Test Prep’s free GRE course. The slideshow is accessed via our SlideShare page ( ) where you can download it to your smartphone/computer for offline viewing. Cheers, Brent – Greenlight Test Prep
  10. If you're still looking for GRE resources, we have a free course with 500+ videos, 600+ practice questions and 5 full-length practice tests. Cheers, Brent - Greenlight Test Prep
  11. Brent Hanneson

    Problem

    You're absolutely correct - nice catch! I have reposted my answer above. Cheers, Brent
  12. Brent Hanneson

    Problem

    Let's examine the two EXTREME values of a. Case a: The GREATEST value of a If the values are strictly increasing (i.e., no values are equal), them a could be a little bit smaller than 8.5. So, a could equal 8.499999999999 For all intents and purposes, let's just say that the a COULD equal 8.5 In this case, 2a = (2)(8.5) = 17, which means the range = 17 - 3.7 = 13.3 Case b: The SMALLEST value of a If the values are strictly increasing (i.e., no values are equal), them 2a could be a little bit bigger than 9.2 So, 2a could equal 9.20000000001 For all intents and purposes, let's just say that the 2a COULD equal 9.2 This mean a = 4.7 This is fine, because we're told that a is greater than 4.1. If 2a = 9.2, then the range = 9.2 - 3.7 = 5.5 Now that we've examined the EXTREME cases, we can see that the range can be between 5.5 and 13.3 Answer(s): C, D and E Cheers, Brent
  13. Please post only one question per thread. Otherwise things can become pretty complicated when there are discussions on multiple questions. Cheers, Brent
  14. Way out of scope for the GRE. Cheers, Brent - Greenlight Test Prep
  15. Hi Yu, You're not alone. Many students find it much easier to follow live (or video) instruction. We happen to have a free GMAT course with over 500 videos. This includes hundreds of practice questions, where the answers are fully explained step-by-step. Cheers, Brent - GMAT Prep Now (see link below)
  16. Funny you should ask, I found this resource on YouTube: It shows images for each GRE word Cheers, Brent - Greenlight Test Prep
  17. With all text completion and sentence equivalence questions, we're looking for descriptive words that provide context for the missing words. People should not be praised for their virtue if they lack the energy to be____ This suggests that people who are too lazy to be NON-VIRTUOUS should not be praised for their virtue. So, the first word must convey the same meaning as NON-VIRTUOUS Hmmm, all of the answer choices look good. So, let's keep going in such cases, goodness is merely the effect of __________. From the first part, we see that their goodness is due to their LAZINESS (there are too lazy to be bad). So, the second word must convey the same meaning as LAZY Only C looks good. When we check both words, they fit nicely. For more strategies on solving text completion questions, see our free set of 51 videos at GRE Text Completion | Greenlight Test Prep Cheers, Brent - Greenlight Test Prep
  18. Analogies are no longer tested on the GRE. Cheers, Brent - Greenlight Test Prep
  19. Since you asked, we have a free video course consisting of over 500 videos, which cover everything you need to ace the GMAT. Cheers, Brent - GMAT Prep Now
  20. I'll echo what bullseye said. Many GRE resources assume a certain amount of pre-existing math skills. Now, there are a lot of great websites out there to learn mathematical concepts (e.g., Khan Academy, Purple Math, etc), HOWEVER none of them address GRE-specific concepts and strategies (quantitative comparison strategies for one). Also, users are often exposed to concepts that are not tested on the GRE, and this can waste what remaining time you have. If possible, I suggest sticking with resources that a specifically towards the GRE. Moreover, find one that starts from the most basic concepts and slowly moves you to 170-level concepts. I believe that our video course is one such resource. Cheers, Brent - Greenlight Test Prep
  21. If you're interested, we have a free GRE prep course consisting of over 500 videos. Cheers, Brent - Greenlight Test Prep
  22. 50 is EVEN, so 50y will be EVEN for any integer y. So, 50y is EVEN and 69 is ODD, which means 50y + 69 is ODD. Since x = 50y + 69, we can conclude that x is ODD. Now check the answer choices to see which one MUST BE ODD: A) xy = (ODD)(y). Since y can be odd or even, this expression can be odd or even - ELIMINATE A B) x+y = ODD + y. Since y can be odd or even, this expression can be odd or even - ELIMINATE B C) x+2y = ODD + 2y. We know that 2y must be EVEN . So , we get x+2y = ODD + EVEN = ODD (perfect!) Answer: C Cheers, Brent
  23. A lot of integer property questions can be solved using prime factorization. For questions involving divisibility, divisors, factors and multiples, we can say: If N is divisible by k, then k is "hiding" within the prime factorization of N Consider these examples: 24 is divisible by 3 because 24 = (2)(2)(2)(3) Likewise, 70 is divisible by 5 because 70 = (2)(5)(7) And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7) And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7) -------------------------------- Okay, onto the question: We want to know how many 6's are hiding in the prime factorization of 252^6 252 = (2)(2)(3)(3)(7) Notice that there are already two 6's hiding in the prime factorization of 252 Since 252^6 = (252)(252)(252)(252)(252)(252), we know that there must be a total of twelve 6's hiding in the prime factorization of 252^6 Answer: E Cheers, Brent
  24. For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is A: Between 2 & 10 B: Between 10 & 20 C: Between 20 & 30 D: Between 30 & 40 E: Greater than 40 Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1 For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1) Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313 Now let’s examine h(100) h(100) = (2)(4)(6)(8)….(96)(98)(100) = (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50) Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)….(48)(49)(50)] Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule) Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule) Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule) . . . . Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule) So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47. Answer = E Cheers, Brent
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